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tech / sci.physics.relativity / Apparent paradox regarding relativistic extension of projectile motion

SubjectAuthor
* Apparent paradox regarding relativistic extension of projectile motionAldo
`* Re: Apparent paradox regarding relativistic extension of projectileTom Roberts
 `* Re: Apparent paradox regarding relativistic extension of projectile motionAldo
  +- Re: Apparent paradox regarding relativistic extension of projectile motionDono.
  `- Re: Apparent paradox regarding relativistic extension of projectile motionAldo

1
Apparent paradox regarding relativistic extension of projectile motion

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Subject: Apparent paradox regarding relativistic extension of projectile motion
From: aldo.may...@cap.edu.mx (Aldo)
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 by: Aldo - Thu, 14 Apr 2022 22:41 UTC

Consider a relativistic charged particle of mass m and charge q subject to a uniform electric field in the y-direction. We can use the relativistic form of Newton's second law F=dp/dt, with p being the relativistic momentum to find
d/dt(γmu)=0, d/dt(γmv)=qE, where:
-u is the x-component of the velocity, u is the y-component of the velocity
-γ=1/√(1-(u²+v²)/c²)
-Initial conditions are u(0)=u_0, v(0)=0

Integrating these equations using the above initial conditions we find that
mu/√(1-(u²+v²)/c²)=p, mv/√(1-(u²+v²)/c²)=qEt, where p=mu_0/√(1-u_0²/c²)
we can now use these equations for u and v as functions of coordinate time, t (but ask a symbolic solver such as Wolfram Alpha).
The result is
u(t)=pc/√(m²c²+p²+q²E²t²), v(t)=cqEt/√(m²c²+p²+q²E²t²),
where the last equation for u(t) calls my attention. There is non-zero acceleration even though there's no force. Why this is so? What's the physical reason?
*DISCLAIMER*: I am not an anti-relativist and this post in no way has the goal to discredit relativity, I'm simply trying to understand it better. In no way I am a Ken Seto or an Ed Lake.

Re: Apparent paradox regarding relativistic extension of projectile motion

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 by: Tom Roberts - Sat, 16 Apr 2022 21:32 UTC

On 4/14/22 5:41 PM, Aldo wrote:
> Consider a relativistic charged particle of mass m and charge q
> subject to a uniform electric field in the y-direction. We can use
> the relativistic form of Newton's second law F=dp/dt, with p being
> the relativistic momentum

(You didn't state it, but you are implicitly assuming
an inertial frame with coordinates {x,y,z,t}. Note also
that all d-s are partial derivatives.)

You confused t, the time coordinate in the inertial frame, with \tau,
the particle's proper time. Using standard notation, the correct
equation for the motion of a pointlike object is:

F = dP/d\tau

where F is the 4-force on the object, P is the object's 4-momentum, and
\tau is the object's proper time. This is manifestly independent of
coordinates (aka invariant), as it must be.

(Around here I write \tau, not τ, because the latter is
too easily confused with t. Also beware of γ vs y.)

> d/dt(γmu)=0, d/dt(γmv)=qE, where:
> -u is the x-component of the velocity, u is the y-component of the velocity

TYPO: v is the y-component of the (3-)velocity.

Hmmm. First replace t => \tau. Then restore the equations you omitted:
d/d\tau(γmw)=0, d/d\tau(γmc^2)=qγEv/c
where w is the z-component of velocity, and γmc^2 is the object's
kinetic energy, both in the inertial frame. These four equations are
coupled, as γ depends on {u,v,w}, each of which is a function of \tau
(in general).

Note that \tau is a function of t and γ; choosing a common origin:
\tau = t/γ
so
d/d\tau = γ d/dt

Using the correct equations will give better results.

Tom Roberts

Re: Apparent paradox regarding relativistic extension of projectile motion

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Subject: Re: Apparent paradox regarding relativistic extension of projectile motion
From: aldo.may...@cap.edu.mx (Aldo)
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 by: Aldo - Sat, 16 Apr 2022 22:36 UTC

El sábado, 16 de abril de 2022 a las 16:33:00 UTC-5, tjrob137 escribió:
> On 4/14/22 5:41 PM, Aldo wrote:
> > Consider a relativistic charged particle of mass m and charge q
> > subject to a uniform electric field in the y-direction. We can use
> > the relativistic form of Newton's second law F=dp/dt, with p being
> > the relativistic momentum
>
> (You didn't state it, but you are implicitly assuming
> an inertial frame with coordinates {x,y,z,t}. Note also
> that all d-s are partial derivatives.)
>
> You confused t, the time coordinate in the inertial frame, with \tau,
> the particle's proper time. Using standard notation, the correct
> equation for the motion of a pointlike object is:
>
> F = dP/d\tau
>
> where F is the 4-force on the object, P is the object's 4-momentum, and
> \tau is the object's proper time. This is manifestly independent of
> coordinates (aka invariant), as it must be.
>
> (Around here I write \tau, not τ, because the latter is
> too easily confused with t. Also beware of γ vs y.)
>
> > d/dt(γmu)=0, d/dt(γmv)=qE, where:
> > -u is the x-component of the velocity, u is the y-component of the velocity
>
> TYPO: v is the y-component of the (3-)velocity.
>
> Hmmm. First replace t => \tau. Then restore the equations you omitted:
> d/d\tau(γmw)=0, d/d\tau(γmc^2)=qγEv/c
> where w is the z-component of velocity, and γmc^2 is the object's
> kinetic energy, both in the inertial frame. These four equations are
> coupled, as γ depends on {u,v,w}, each of which is a function of \tau
> (in general).
>
> Note that \tau is a function of t and γ; choosing a common origin:
> \tau = t/γ
> so
> d/d\tau = γ d/dt
>
> Using the correct equations will give better results.
>
> Tom Roberts
Thank you Tom, although you should note that I was doing my calculations using a "relativistic but non covariant" formulation of SR dynamics, using the coordinate time t as the evolution parameter, and using ordinary, Euclidean 3-vectors instead of 4-vectors, and using the corresponding relativistic expressions for momentum, p=γmv, and energy, E=γmc², with the "relativistic but noncovariant" Newton's second law, F=d/dt(γmv).
The answer to the question was actually very simple, in SR, the 3-vectors F, the force, and a, the acceleration no longer have the same direction. Indeed one can see by simple differentiation using the product rule that F=d/dt(γmv)=γma+(m/c²)(v∙a)v, and from this it is obvious that F and a no longer have the same direction.

The "paradox" would immediately disappear if we were to use an explicitly relativistic formulation, with the equation of motion being dP/dτ=qGU, with now P being the four momentum, U the four velocity and G the electronagnetic field tensor, then setting the electric field to be a constant in the y-direction and zero in the x and z directions, and yields x(τ), y(τ) and t(τ)

Re: Apparent paradox regarding relativistic extension of projectile motion

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Subject: Re: Apparent paradox regarding relativistic extension of projectile motion
From: eggy2001...@gmail.com (Dono.)
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 by: Dono. - Sat, 16 Apr 2022 22:53 UTC

On Saturday, April 16, 2022 at 3:36:28 PM UTC-7, Aldo wrote:
> El sábado, 16 de abril de 2022 a las 16:33:00 UTC-5, tjrob137 escribió:
> > On 4/14/22 5:41 PM, Aldo wrote:
> > > Consider a relativistic charged particle of mass m and charge q
> > > subject to a uniform electric field in the y-direction. We can use
> > > the relativistic form of Newton's second law F=dp/dt, with p being
> > > the relativistic momentum
> >
> > (You didn't state it, but you are implicitly assuming
> > an inertial frame with coordinates {x,y,z,t}. Note also
> > that all d-s are partial derivatives.)
> >
> > You confused t, the time coordinate in the inertial frame, with \tau,
> > the particle's proper time. Using standard notation, the correct
> > equation for the motion of a pointlike object is:
> >
> > F = dP/d\tau
> >
> > where F is the 4-force on the object, P is the object's 4-momentum, and
> > \tau is the object's proper time. This is manifestly independent of
> > coordinates (aka invariant), as it must be.
> >
> > (Around here I write \tau, not τ, because the latter is
> > too easily confused with t. Also beware of γ vs y.)
> >
> > > d/dt(γmu)=0, d/dt(γmv)=qE, where:
> > > -u is the x-component of the velocity, u is the y-component of the velocity
> >
> > TYPO: v is the y-component of the (3-)velocity.
> >
> > Hmmm. First replace t => \tau. Then restore the equations you omitted:
> > d/d\tau(γmw)=0, d/d\tau(γmc^2)=qγEv/c
> > where w is the z-component of velocity, and γmc^2 is the object's
> > kinetic energy, both in the inertial frame. These four equations are
> > coupled, as γ depends on {u,v,w}, each of which is a function of \tau
> > (in general).
> >
> > Note that \tau is a function of t and γ; choosing a common origin:
> > \tau = t/γ
> > so
> > d/d\tau = γ d/dt
> >
> > Using the correct equations will give better results.
> >
> > Tom Roberts
> Thank you Tom, although you should note that I was doing my calculations using a "relativistic but non covariant" formulation of SR dynamics, using the coordinate time t as the evolution parameter, and using ordinary, Euclidean 3-vectors instead of 4-vectors, and using the corresponding relativistic expressions for momentum, p=γmv, and energy, E=γmc², with the "relativistic but noncovariant" Newton's second law, F=d/dt(γmv).
> The answer to the question was actually very simple, in SR, the 3-vectors F, the force, and a, the acceleration no longer have the same direction. Indeed one can see by simple differentiation using the product rule that F=d/dt(γmv)=γma+(m/c²)(v∙a)v, and from this it is obvious that F and a no longer have the same direction.
>
> The "paradox" would immediately disappear if we were to use an explicitly relativistic formulation, with the equation of motion being dP/dτ=qGU, with now P being the four momentum, U the four velocity and G the electronagnetic field tensor, then setting the electric field to be a constant in the y-direction and zero in the x and z directions, and yields x(τ), y(τ) and t(τ)

The covariant formulation produces the more elegant results but the coordinate-dependent formulation that you correctly used in your original post produces THE useful results since we are interested in the equations of motion wrt the lab reference frame. As I pointed out to you , it is perfectly normal that coordinate acceleration is not colinear with coordinate force. No paradox.

Re: Apparent paradox regarding relativistic extension of projectile motion

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Subject: Re: Apparent paradox regarding relativistic extension of projectile motion
From: aldo.may...@cap.edu.mx (Aldo)
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 by: Aldo - Sat, 16 Apr 2022 22:57 UTC

El sábado, 16 de abril de 2022 a las 17:36:28 UTC-5, Aldo escribió:
> El sábado, 16 de abril de 2022 a las 16:33:00 UTC-5, tjrob137 escribió:
> > On 4/14/22 5:41 PM, Aldo wrote:
> > > Consider a relativistic charged particle of mass m and charge q
> > > subject to a uniform electric field in the y-direction. We can use
> > > the relativistic form of Newton's second law F=dp/dt, with p being
> > > the relativistic momentum
> >
> > (You didn't state it, but you are implicitly assuming
> > an inertial frame with coordinates {x,y,z,t}. Note also
> > that all d-s are partial derivatives.)
> >
> > You confused t, the time coordinate in the inertial frame, with \tau,
> > the particle's proper time. Using standard notation, the correct
> > equation for the motion of a pointlike object is:
> >
> > F = dP/d\tau
> >
> > where F is the 4-force on the object, P is the object's 4-momentum, and
> > \tau is the object's proper time. This is manifestly independent of
> > coordinates (aka invariant), as it must be.
> >
> > (Around here I write \tau, not τ, because the latter is
> > too easily confused with t. Also beware of γ vs y.)
> >
> > > d/dt(γmu)=0, d/dt(γmv)=qE, where:
> > > -u is the x-component of the velocity, u is the y-component of the velocity
> >
> > TYPO: v is the y-component of the (3-)velocity.
> >
> > Hmmm. First replace t => \tau. Then restore the equations you omitted:
> > d/d\tau(γmw)=0, d/d\tau(γmc^2)=qγEv/c
> > where w is the z-component of velocity, and γmc^2 is the object's
> > kinetic energy, both in the inertial frame. These four equations are
> > coupled, as γ depends on {u,v,w}, each of which is a function of \tau
> > (in general).
> >
> > Note that \tau is a function of t and γ; choosing a common origin:
> > \tau = t/γ
> > so
> > d/d\tau = γ d/dt
> >
> > Using the correct equations will give better results.
> >
> > Tom Roberts
> Thank you Tom, although you should note that I was doing my calculations using a "relativistic but non covariant" formulation of SR dynamics, using the coordinate time t as the evolution parameter, and using ordinary, Euclidean 3-vectors instead of 4-vectors, and using the corresponding relativistic expressions for momentum, p=γmv, and energy, E=γmc², with the "relativistic but noncovariant" Newton's second law, F=d/dt(γmv).
> The answer to the question was actually very simple, in SR, the 3-vectors F, the force, and a, the acceleration no longer have the same direction. Indeed one can see by simple differentiation using the product rule that F=d/dt(γmv)=γma+(m/c²)(v∙a)v, and from this it is obvious that F and a no longer have the same direction.
>
> The "paradox" would immediately disappear if we were to use an explicitly relativistic formulation, with the equation of motion being dP/dτ=qGU, with now P being the four momentum, U the four velocity and G the electronagnetic field tensor, then setting the electric field to be a constant in the y-direction and zero in the x and z directions, and yields x(τ), y(τ) and t(τ)
*Typing mistake with the SR Newton's second law, it should read F=d/dt(γmv)=γma+(mγ³/c²)(v∙a)v

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